Question: Divide the following complex numbers. $ \dfrac{-20+12i}{3+5i}$
Solution: We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${3-5i}$ $ \dfrac{-20+12i}{3+5i} = \dfrac{-20+12i}{3+5i} \cdot \dfrac{{3-5i}}{{3-5i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(-20+12i) \cdot (3-5i)} {(3+5i) \cdot (3-5i)} = \dfrac{(-20+12i) \cdot (3-5i)} {3^2 - (5i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(-20+12i) \cdot (3-5i)} {(3)^2 - (5i)^2} = $ $ \dfrac{(-20+12i) \cdot (3-5i)} {9 + 25} = $ $ \dfrac{(-20+12i) \cdot (3-5i)} {34} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({-20+12i}) \cdot ({3-5i})} {34} = $ $ \dfrac{{-20} \cdot {3} + {12} \cdot {3 i} + {-20} \cdot {-5 i} + {12} \cdot {-5 i^2}} {34} $ Evaluate each product of two numbers. $ \dfrac{-60 + 36i + 100i - 60 i^2} {34} $ Finally, simplify the fraction. $ \dfrac{-60 + 36i + 100i + 60} {34} = \dfrac{0 + 136i} {34} = 4i $